Leetcode 1319: Number of Operations to Make Network Connected

There are n computers numbered from 0 to n - 1 connected by ethernet cables connections forming a network where connections[i] = [ai, bi] represents a connection between computers ai and bi. Any computer can reach any other computer directly or indirectly through the network.

You are given an initial computer network connections. You can extract certain cables between two directly connected computers, and place them between any pair of disconnected computers to make them directly connected.

Return the minimum number of times you need to do this in order to make all the computers connected. If it is not possible, return -1.

Example 1:

Input: n = 4, connections = [[0,1],[0,2],[1,2]]
Output: 1
Explanation: Remove cable between computer 1 and 2 and place between computers 1 and 3.

Example 2:

Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2],[1,3]]
Output: 2

Example 3:

Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2]]
Output: -1
Explanation: There are not enough cables.

Constraints:

• 1 <= n <= 10^5
• 1 <= connections.length <= min(n * (n - 1) / 2, 10^5)
• connections[i].length == 2
• 0 <= ai, bi < n
• ai != bi
• There are no repeated connections.
• No two computers are connected by more than one cable.

Problem Analysis:

We first need to find the number of subnetworks, that every computers in the same subnetwork is connected with each other directly or indirectly.

Then we need to find redundant cables which when we remove the cable it will not change subnetwork status.

Assume the number of subnetworks is nets, and the number of redundant cables is cables, if cables ≥ nets-1, then we can make all subnetworks connected, which means all computers can connect directly or indirectly.

So we can use Union-Find algorithm to get the number of subnetworks and the number of redundant cables, and then find the answer.

Solution

class Solution {
public:
    int makeConnected(int n, vector<vector<int>>& connections) {
        parents_.resize(n);
        rank_.resize(n);
        iota(begin(parents_), end(parents_), 0);
        networks_ = n;
        
        for (const auto& item: connections) {
            int x = find(item.front());
            int y = find(item.back());
            if (x == y) {
                ++redundant_;
            } else {
                --networks_;
                unite(item.front(), item.back());
            }
        }

        return (redundant_ < networks_ - 1) ? -1 : (networks_ - 1);
    }

    int find(int x) {
        if (x == parents_[x]) return x;
        return parents_[x] = find(parents_[x]);
    }

    void unite(int x, int y) {
        x = find(x);
        y = find(y);

        if (x != y) {
            if (rank_[x] > rank_[y]) {
                parents_[y] = x;
            } else {
                if (rank_[x] == rank_[y]) ++rank_[y];
                parents_[x] = y;
            }
        }
    }

private:
    vector<int> parents_;
    vector<int> rank_;
    int networks_;
    int redundant_{0};
};

Time complexity is O(n)

Space complexity is O(n)